let U_n = ∫_0 ^∞ ((cos(nx))/((x^2 +n^2 )^3 ))dx with n≥1 1) calculate U_n intrems of n 2) find lim_(n→+∞) n U_n 3) calculate lim_(n→+∞) n^2 U_n 4) study the convervence of U


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Question Number 62128 by maxmathsup by imad last updated on 15/Jun/19

$l e t U_{n} = \int_{0}^{\infty} \frac{c o s (n x)}{{(x^{2} + n^{2})}^{3}} d x w i t h n ⩾ 1$ $1) c a l c u l a t e U_{n} i n t r e m s o f n$ $2) f i n d {l i m}_{n \to + \infty} n U_{n}$ $3) c a l c u l a t e {l i m}_{n \to + \infty} n^{2} U_{n}$ $4) s t u d y t h e c o n v e r v e n c e o f U_{n}$
Commented by maxmathsup by imad last updated on 16/Jun/19
$1) U_n =∫_0 ^∞ ((cos(nx))/((x^2 +n^2 )^3 ))dx changement x =nt give U_n =∫_0 ^∞ ((cos(n^2 t))/((n^2 t^2 +n^2 )^3 )) ndt =(1/n^5 ) ∫_0 ^∞ ((cos(n^2 t))/((t^2 +1)^3 )) dt ⇒2n^5 U_n =∫_(−∞) ^(+∞) ((cos(n^2 t))/((t^2 +1)^3 ))dt =Re(∫_(−∞) ^(+∞) (e^(in^2 t) /((t^2 +1)^3 ))dt) let consider thecomplex function ϕ(z)=(e^(in^2 z) /((z^2 +1)^3 )) we have ϕ(z)=(e^(inz^2 ) /((z−i)^3 (z+i)^3 )) the poles of ϕ are +^− i (triples) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) Res(ϕ,i) =lim_(z→i) (1/((3−1)!)){(z−i)^3 ϕ(z)}^((2)) =lim_(z→i) (1/2){ (e^(in^2 z) /((z+i)^3 ))}^((2)) but { (e^(in^2 z) /((z+i)^3 ))}^((2)) ={((in^2 e^(in^2 z) (z+i)^3 −3(z+i)^2 e^(in^2 z) )/((z+i)^6 ))}^((1)) ={(((in^2 (z+i)−3)e^(in^2 z) )/((z+i)^4 ))}^((1)) ={(((in^2 z−n^2 −3)e^(in^2 z) )/((z+i)^4 ))}^((1)) =(((in^2 +in^2 (in^2 z−n^2 −3))e^(in^2 z) (z+i)^4 −4(z+i)^3 (in^2 z−n^2 −3)e^(in^2 z) )/((z+i)^8 )) =(({(in^2 −n^4 z −in^4 −3in^2 )(z+i)−4in^2 z+4n^2 +12}e^(in^2 z) )/((z+i)^5 )) ⇒ Res(ϕ,i) =(1/2) (({ 2i(in^2 −n^4 i−in^4 −3in^2 ) +4n^2 +4n^2 +12}e^(−n^2 ) )/((2i)^5 )) =(1/2^6 ) (({−2n^2 +2n^4 +2n^4 +6n^2 +8n^2 +12}e^(−n^2 ) )/i) =(1/(i 2^6 )){12n^2 +4n^4 +12}e^(−n^2 ) =(4/(i 2^6 )){ 3n^2 +n^4 +3}e^(−n^2 ) = (1/(16i)){n^4 +3n^2 +3} e^(−n^2 ) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =((2iπ)/(16i)){ n^4 +3n^2 +3}e^(−n^2 ) =(π/8)(n^4 +3n^2 +3)e^(−n^2 ) =2n^5 U_n ⇒ U_n =(π/(16n^5 ))(n^4 +3n^2 +3)e^(−n^2 ) 3) n^2 U_n =(π/(16n^3 ))(n^4 +3n^2 +3)e^(−n^2 ) =(π/(16))(n +3 +(3/n^3 ))e^(−n^2 ) ⇒ lim_(n→+∞) n^2 U_n =lim_(n→+∞) (π/(16))(n+3)e^(−n^2 ) =0$
$1) U_{n} = \int_{0}^{\infty} \frac{c o s (n x)}{{(x^{2} + n^{2})}^{3}} d x c h a n g e m e n t x = n t g i v e$ $U_{n} = \int_{0}^{\infty} \frac{c o s (n^{2} t)}{{(n^{2} t^{2} + n^{2})}^{3}} n d t = \frac{1}{n^{5}} \int_{0}^{\infty} \frac{c o s (n^{2} t)}{{(t^{2} + 1)}^{3}} d t \Rightarrow 2 n^{5} U_{n} = \int_{- \infty}^{+ \infty} \frac{c o s (n^{2} t)}{{(t^{2} + 1)}^{3}} d t$ $= R e (\int_{- \infty}^{+ \infty} \frac{e^{{i n}^{2} t}}{{(t^{2} + 1)}^{3}} d t) l e t c o n s i d e r t h e c o m p l e x f u n c t i o n φ (z) = \frac{e^{{i n}^{2} z}}{{(z^{2} + 1)}^{3}}$ $w e h a v e φ (z) = \frac{e^{{i n z}^{2}}}{{(z - i)}^{3} {(z + i)}^{3}} t h e p o l e s o f φ a r e \bar{+} i (t r i p l e s) r e s i d u s t h e o r e m$ $g i v e \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i)$ $R e s (φ, i) = {l i m}_{z \to i} \frac{1}{(3 - 1)!} {{(z - i)}^{3} φ (z)}^{(2)}$ $= {l i m}_{z \to i} \frac{1}{2} {\frac{e^{{i n}^{2} z}}{{(z + i)}^{3}}}^{(2)} b u t$ ${\frac{e^{{i n}^{2} z}}{{(z + i)}^{3}}}^{(2)} = {\frac{{i n}^{2} e^{{i n}^{2} z} {(z + i)}^{3} - 3 {(z + i)}^{2} e^{{i n}^{2} z}}{{(z + i)}^{6}}}^{(1)}$ $= {\frac{({i n}^{2} (z + i) - 3) e^{{i n}^{2} z}}{{(z + i)}^{4}}}^{(1)} = {\frac{({i n}^{2} z - n^{2} - 3) e^{{i n}^{2} z}}{{(z + i)}^{4}}}^{(1)}$ $= \frac{({i n}^{2} + {i n}^{2} ({i n}^{2} z - n^{2} - 3)) e^{{i n}^{2} z} {(z + i)}^{4} - 4 {(z + i)}^{3} ({i n}^{2} z - n^{2} - 3) e^{{i n}^{2} z}}{{(z + i)}^{8}}$ $= \frac{{({i n}^{2} - n^{4} z - {i n}^{4} - 3 {i n}^{2}) (z + i) - 4 {i n}^{2} z + 4 n^{2} + 12} e^{{i n}^{2} z}}{{(z + i)}^{5}} \Rightarrow$ $R e s (φ, i) = \frac{1}{2} \frac{{2 i ({i n}^{2} - n^{4} i - {i n}^{4} - 3 {i n}^{2}) + 4 n^{2} + 4 n^{2} + 12} e^{- n^{2}}}{{(2 i)}^{5}}$ $= \frac{1}{2^{6}} \frac{{- 2 n^{2} + 2 n^{4} + 2 n^{4} + 6 n^{2} + 8 n^{2} + 12} e^{- n^{2}}}{i} = \frac{1}{i 2^{6}} {12 n^{2} + 4 n^{4} + 12} e^{- n^{2}}$ $= \frac{4}{i 2^{6}} {3 n^{2} + n^{4} + 3} e^{- n^{2}} = \frac{1}{16 i} {n^{4} + 3 n^{2} + 3} e^{- n^{2}} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) d z = \frac{2 i π}{16 i} {n^{4} + 3 n^{2} + 3} e^{- n^{2}} = \frac{π}{8} (n^{4} + 3 n^{2} + 3) e^{- n^{2}} = 2 n^{5} U_{n} \Rightarrow$ $U_{n} = \frac{π}{16 n^{5}} (n^{4} + 3 n^{2} + 3) e^{- n^{2}}$ $3) n^{2} U_{n} = \frac{π}{16 n^{3}} (n^{4} + 3 n^{2} + 3) e^{- n^{2}} = \frac{π}{16} (n + 3 + \frac{3}{n^{3}}) e^{- n^{2}} \Rightarrow$ ${l i m}_{n \to + \infty} n^{2} U_{n} = {l i m}_{n \to + \infty} \frac{π}{16} (n + 3) e^{- n^{2}} = 0$
Commented by maxmathsup by imad last updated on 16/Jun/19

$2) w e h a v e {n U}_{n} = \frac{π}{16 n^{4}} (n^{4} + 3 n^{2} + 3) e^{- n^{2}} \Rightarrow {l i m}_{n \to + \infty} {n U}_{n} =$ ${l i m}_{n \to + \infty} \frac{π}{16} e^{- n^{2}} = 0$
Commented by maxmathsup by imad last updated on 16/Jun/19

$4) t h e Q i s s t u d y t h e c o n v e r g e n c e o f Σ U_{n}$
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